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CMC Medical CMC-Medical VELLORE Solved Paper-2011

  • question_answer
    The magnitude of the force developed by raising the temperature from \[0{}^\circ C\] to \[100{}^\circ C\] of the iron bar 1.0 m long and \[1c{{m}^{2}}\]cross-section, when it is held so that it is not permitted to expand or bend is \[\left( \alpha ={{10}^{5}}{{/}^{0}}C,Y={{10}^{11}}N/{{m}^{2}} \right)\]

    A)  \[{{10}^{3}}N\]                                

    B)  \[{{10}^{4}}N\]

    C)  \[{{10}^{5}}N\]                

    D)  \[{{10}^{9}}N\]

    E)  \[{{10}^{2}}N\]

    Correct Answer: B

    Solution :

                    \[F=YA\frac{\Delta L}{L}=\frac{YA}{L}\times \alpha L\Delta \theta =YA\,\alpha \Delta \theta \] \[\therefore \]\[F={{10}^{11}}\times {{10}^{-4}}\times {{10}^{-5}}\times 100\] \[={{10}^{4}}N\]


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